In the following equation X is: ₉₅²⁴¹ Am+α → ₉₇²⁴³ Bk+X · ABC26IN0738

Subject: Inorganic Chemistry · Chapter: Nuclear Chemistry · Topic: Nuclear Chemistry – General · Exam: JAM 2005-2022 · Marks: 2 · Difficulty: Easy

In the following equation X is:

\[ { }_{95}^{241} \mathrm{Am}+\alpha \longrightarrow{ }_{97}^{243} \mathrm{Bk}+\mathrm{X} \]

(a)$2{ }_{0}^{1} \mathrm{n}$

(b)${ }_{0}^{1} \mathrm{n}$

(c)$2{ }_{1}^{1} \mathrm{n}$

(d)${ }_{2}^{4} \mathrm{He}$

Answer

Answer: A ✓ checked by 4AB · confidence high

Explanation

²⁴¹Am + ⁴He → ²⁴³Bk + 2 ¹n: the bombarding α raises Z by 2 and the mass by 4, and two neutrons are released to leave A = 243.

Study loop for Nuclear Chemistry